\newproblem{lay:5_5_13}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.5.13}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let the matrix $A=\begin{pmatrix}1 & -2 \\ 1 &3\end{pmatrix}$. Find an invertible matrix $P$ and a matrix $C$ of the form $\begin{pmatrix} b & -a \\ a & b\end{pmatrix}$ such that
	$A=PCP^{-1}$.
}{
  % Solution
	As in Exercise 5.5.1, the eigenvalues of $A$ are $\lambda=2\pm i$. A basis of the eigenvalue $\lambda=2-i$ is $\{(-1-i),1\}$. According to Theorem 5.5.9, $\lambda=a-bi$ and
	$\mathbf{v}$ is a basis of its eigenspace, we find the $P$ and $C$ matrices as
	\begin{center}
		$P=\begin{pmatrix} \mathrm{Re}\{\mathbf{v}\} & \mathrm{Im}\{\mathbf{v}\} \end{pmatrix}$\\
		$C=\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$
	\end{center}
	In this case,
	\begin{center}
		$P=\begin{pmatrix} -1 & -1 \\ 1 & 0 \end{pmatrix}$\\
		$C=\begin{pmatrix} 2 & -1 \\ 1 & 2 \end{pmatrix}$
	\end{center}
	It can be easily verified that
	\begin{center}
		$A=\begin{pmatrix}1 & -2 \\ 1 & 3\end{pmatrix}=\begin{pmatrix} -1 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 2 & -1 \\ 1 & 2 \end{pmatrix}
		   \begin{pmatrix} -1 & -1 \\ 1 & 0 \end{pmatrix}^{-1}$
	\end{center}
}
\useproblem{lay:5_5_13}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
